Wave Motion Ques 62
- A string of length $1 m$ and mass $5 g$ is fixed at both ends. The tension in the string is $8.0 N$. The string is set into vibration using an external vibrator of frequency $100 Hz$. The separation between successive nodes on the string is close to
(2019 Main, 10 Jan I)
(a) $16.6 cm$
(b) $33.3 cm$
(c) $10.0 cm$
(d) $20.0 cm$
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Answer:
Correct Answer: 62.(d)
Solution:
Formula:
- Velocity ’ $v$ ’ of the wave on the string $=\sqrt{\frac{T}{\mu}}$
where, $T=$ tension and $\mu=$ mass per unit length.
Substituting the given values, we get
$$ v=\sqrt{\frac{8}{5} \times 1000}=40 ms^{-1} $$
Wavelength of the wave on the string, $\lambda=\frac{v}{f}$
where, $\quad f=$ frequency of wave.
$$ \Rightarrow \quad \lambda=\frac{40}{100} m=40 cm $$
$\therefore$ Separation between two successive nodes is,
$$ d=\frac{\lambda}{2}=\frac{40}{2}=20.0 cm $$
BETA
